29
$\begingroup$

So this may be one of the stupidest questions ever, but is the identity

$$1 + 1 = 2$$

valid in any base?

It seems so, since it's like, for a given base $b$

$$1 = 000\ldots 01 = (0\cdot b^n) + \ldots (1\cdot b^0) = b^0 = 1$$

Which means $1 + 1 = 2$ no matter what base we use.

Is that right? Or is there something I miss?

$\endgroup$
  • 72
    $\begingroup$ There are 10 different kinds of people in the world: those who understand binary, and those who do not. $\endgroup$ – Alex Ortiz Aug 18 '16 at 18:54
  • 20
    $\begingroup$ Well, it's a syntax error in binary if that counts. $\endgroup$ – ApproachingDarknessFish Aug 18 '16 at 19:03
  • 10
    $\begingroup$ @AOrtiz Right now your comment has $10$ upvotes. Two more and you'll have $100$. $\endgroup$ – John Aug 18 '16 at 19:48
  • 48
    $\begingroup$ Actually, there are 10 types of people in the world: those who understand binary, those who don't, and those who weren't expecting a ternary joke. $\endgroup$ – Charlie Hills Aug 18 '16 at 20:30
  • 5
    $\begingroup$ @CharlieHills, I feel like I've had this conversation before. Extensively. $\endgroup$ – Wildcard Aug 18 '16 at 20:48
45
$\begingroup$

Yes, this is valid in any base in which $1$ and $2$ are both digits (so, with the standard conventions, any base except base $2$). More generally, a single digit always represents the same number no matter what base you consider it in (as long as it is a valid digit in that base). So for instance, $3+4=7$ is valid when interpreted in any base (as long as the base is at least $8$, so these are all digits in the base).

To be more precise, we should be clear to distinguish numbers from the sequences of digits we might use to represent them. Standard notation unfortunately does not make this very clear. When we write $1+1=2$ normally, what we really mean is "the sum of the number represented by $1$ in decimal notation and the number represented by $1$ in decimal notation is the number represented by $2$ in decimal notation". So what "$1+1=2$ is valid in any base" really means is "for any base $b>2$, the sum of the number represented by $1$ in base $b$ notation and the number represented by $1$ in base $b$ notation is the number represented by $2$ in base $b$ notation." This is because, as mentioned above, "the number represented by $1$ in base $b$ notation" is the exact same number as "the number represented by $1$ in decimal notation", and similarly for $2$.

$\endgroup$
  • 7
    $\begingroup$ I would go a bit further and say that these equations still make sense (and are still true) if you allow numbers beyond the base (which is done from time to time, e.g. in discussions of carrying). Then $1+1=2$ is valid in all bases, even binary and even unary. $\endgroup$ – Mario Carneiro Aug 18 '16 at 19:03
  • 4
    $\begingroup$ I feel poorly for unary which is left out of so many discussion of base just because it's so often useless. It should be included with binary as a base in which $1+1=2$ is invalid. $1+1=11$ $\endgroup$ – Engineer Toast Aug 18 '16 at 21:14
  • $\begingroup$ It gets fun when you look at how the sum of digits of multiples of n - 1 in base n are multiples of n - 1. $\endgroup$ – zzzzBov Aug 19 '16 at 3:46
  • $\begingroup$ @EngineerToast In binary, $1+1=10$ $\endgroup$ – rexkogitans Aug 19 '16 at 6:44
  • 1
    $\begingroup$ @rexkogitans He's speaking about unary. $\endgroup$ – bezmax Aug 19 '16 at 6:53
25
$\begingroup$

Integers have their own existence, separate from how we may choose to represent them.

The statement

"(the integer referred to by the decimal symbol 1) + (the integer referred to by the decimal symbol 1) = (the integer referred to by the decimal symbol 2)"

is indeed always true. Whether or not this is expressed in symbols as

"1 + 1 = 2"

depends on how you choose to represent integers.

$\endgroup$
  • 3
    $\begingroup$ IMHO, you're answering a different question. This question, since it mentions "in any base", is asking about ways of representing integers. The statement "(the integer referred to by the decimal symbol 5) + (the integer referred to by the decimal symbol 5) = (the integer referred to by the decimal symbol 10)" is also always true, but so what? "5+5=10" is true only in decimal. $\endgroup$ – Scott Aug 18 '16 at 19:56
  • 10
    $\begingroup$ IMHO, it makes no sense to say "'5+5=10' is true only in decimal". A mathematical statement about integers is true or false. You can then proceed to say that a certain sequence of symbols is a valid (or invalid) representation of that statement in a given system. I would say "the string 5+5=10 is a valid representation of the mathematical fact that (the integer 5) + (the integer 5) = (the integer 10) in the decimal system of representation, or any system where 5 means (the integer 5) and 10 means (the integer 10)". $\endgroup$ – Zev Chonoles Aug 18 '16 at 20:12
  • 4
    $\begingroup$ However, I do grant the statement that "the string 5+5=10 is a valid representation of a true mathematical statement in the base b system of representation if and only if b is (the integer 10)", which perhaps is what you were expressing. $\endgroup$ – Zev Chonoles Aug 18 '16 at 20:17
  • 3
    $\begingroup$ "5 + 5 = 10" only means "5+5 = 10" in decimal. In Base 8 it means "5+ 5 = 8" which is false and in hexidecimal it means 5 + 5 = 16. In base 5 or lower it is meaningless gibberish as "5" doesn't mean anything."'5+5=10' is true only in decimal" makes utterly perfect sense to me as "5+5 = 10" only represents a true statement in decimal. The "'5+5 =10'" was put in quotes. So it's like saying "'poison' is fish only in french" To argue that poison is always poison and only fish when it's fugu bladder utterly misses the point. $\endgroup$ – fleablood Aug 18 '16 at 23:41
  • 1
    $\begingroup$ @flea, the French word is "poisson" (same spelling as that mathematician, if it helps). This undermines your assertion a wee bit. $\endgroup$ – J. M. is a poor mathematician Aug 19 '16 at 8:54
15
$\begingroup$

That's a good question! According to the rules of arithmetic, 1 + 1 = 2 is a true statement about numbers.

Therefore, any reasonable way of representing numbers—whether base 2 or base 16 or base 10— should allow you to express that true statement.

On the other hand, maybe you're asking about symbols rather than numbers: maybe you're asking whether the symbolic expression 1+1 = 2 is true in any base $b$ (where we interpret 1 and 2 as symbols in base $b$).

The answer is yes: the statement is true in any base $b>2$. The reason is that for any base $b>2$, the symbol 2 is a meaningful symbol in base $b$; it refers to $2\cdot b^0$. And we have that $b^0 + b^0 = 2\cdot b^0$ by arithmetic— hence the symbolic expression 1 + 1 = 2 is true in any base $b>2$.

$\endgroup$
8
$\begingroup$

It is always valid because $1+1$ is the definition of $2$. It's not a theorem, it's what we mean by the symbol $2$ to begin with.

$\endgroup$
  • $\begingroup$ Well strictly speaking it's what you mean by a base $>2$. In base $2$ the symbol $2$ is (usually) not defined. $\endgroup$ – Cronus Aug 18 '16 at 20:19
  • $\begingroup$ It's not good to define a symbol in terms of other symbols. That's how you convey the definition, but don't confuse the symbols that represent the definition with the actual meaning of the symbols. There are other ways to convey a definition. (Blocks of wood, for example.) $\endgroup$ – Wildcard Aug 18 '16 at 21:00
  • $\begingroup$ We don't define a symbol in terms of other symbols we're defining a concept in terms of other concepts and simply using symbols to represent the concepts. $\endgroup$ – Gregory Grant Aug 18 '16 at 21:12
5
$\begingroup$

There are three forces at play here.

First comes our natural understanding of numbers. $2$ is a specific natural, or real, number. We understand it intuitively, and unambiguously. This is the abstraction of the idea of having two apples, two sons, two cats, or two examples for collections with two objects. As such, when we think of the number $2$, it is literally defined to be $1+1$.

Then comes the representation of a number in a particular base. This is now a question of how we dress the abstract number into a slightly more tangible form. Here $2$ is a relatively "bad" example, as most [natural-]bases are large enough so $2$ is just $2$ again. But think of the number ten. In decimal, this is $10$; in trenary this is $101$; in octal, $12$. All of these are different strings which represent the same number. How is this possible? Well, when we change base, we change how we interpret the string of symbols.

And finally comes the purely syntactical evaluation of symbols. This takes the last point and pushes it to $11$.(1) We forget that the symbols even represent the abstract quantity of how many hands a "standard" person should have. We only know that $1$, $2$ and $+$ are symbols in a mathematical language. As such we are in fact allowed to interpret them to mean pretty much anything that we want. You see this with $\pi$, which can denote a homomorphism, a projection, a function, a constant real number; but you see it less often with $2$ or $+$. Nevertheless, we are allowed to interpret those symbols to our liking, and we can easily concoct interpretations where $1+1$ and $2$ are not the same object.

In short, if you consider them as abstract numbers, they are independent of their representations, and then $1+1=2$, always. If you are asking about the interpretation of the symbols as numbers, then the answer is a qualified "usually". If you are asking in general about interpreting the symbols $1,2$ and $+$... then this is a resounding "not enough context".


(1) This can be taken in any base you prefer. Even in base $\omega$.

$\endgroup$
  • $\begingroup$ Very, very good answer but I think it doesn't answer if 14+13= 27 in base ten but 14 + 13 = 32 in base 5, does 1 + 1 = 2 in all bases. Then again I think all answers, including mine, elided it. $\endgroup$ – fleablood Aug 19 '16 at 17:42
  • $\begingroup$ The point that I am trying to make is that you have three ways to read this. If you are talking about the abstract numbers, then the answer is positive. If the meaning is the string being interpreted as numbers in various representations, the answer is "not really", and if we ignore the context of numbers, then we get a resounding no. $\endgroup$ – Asaf Karagila Aug 19 '16 at 17:46
4
$\begingroup$

No. There are lots of bases in which your expression does not make sense, e.g.

  • Base 1

    ┌───────────┬───────────┐
    │    Base 1 │   Base 10 │
    ├───────────┼───────────┤
    │         1 │         1 │
    │        11 │         2 │
    │       111 │         3 │
    │      1111 │         4 │
    │       ... │       ... │
    │    1+1=11 │     1+1=2 │
    │     1+1=2 │     Error │ Illegal character "2"
    └───────────┴───────────┘
    
  • Base 2

    ┌───────────┬───────────┐
    │    Base 2 │   Base 10 │
    ├───────────┼───────────┤
    │         1 │         1 │
    │        10 │         2 │
    │        11 │         3 │
    │       100 │         4 │
    │       ... │       ... │
    │    1+1=10 │     1+1=2 │
    │     1+1=2 │     Error │ Illegal character "2"
    └───────────┴───────────┘
    
  • Base φ in standard form (non-unique)

    ┌───────────┬───────────┐
    │    Base φ │   Base 10 │
    ├───────────┼───────────┤
    │         1 │         1 │
    │     10.01 │         2 │
    │    100.01 │         3 │
    │    101.01 │         4 │
    │       ... │       ... │
    │ 1+1=10.01 │     1+1=2 │
    │     1+1=2 │     Error │ Illegal character "2"
    └───────────┴───────────┘
    

Even when restricting to bases which contain the digit "2", it's still not true.

Digits are only representations, and they can represent any value. We usually use base 10 with the digits "0", "1", "2", "3", "4", "5", "6", "7", "8"and "9". So when working with other bases, we reuse these digits if possible, or we borrow latin alphabet letters if we need more, in order to avoid confusion.

But I can define a base 5 positional system with the digits "0", "2", "$", "1", "€". Let's call it "Oriol's system".

┌───────────┬───────────┐
│     Oriol │   Base 10 │
├───────────┼───────────┤
│         2 │         1 │
│         $ │         2 │
│         1 │         3 │
│         € │         4 │
│       ... │       ... │
│     2+2=$ │     1+1=2 │
│     1+1=2 │     3+3=1 │
└───────────┴───────────┘

In that system, the expression 1+1=2 makes sense, but it's false.

Your statement is only true in positional bases whose representations contain the digits "1" and "2", and their meaning is the same as in the decimal system.

$\endgroup$
  • 1
    $\begingroup$ Your view of a "base" is way too narrow. I can, for one, use the digit two in the binary base. Who has forbid it?! $\endgroup$ – yo' Aug 18 '16 at 22:32
  • 3
    $\begingroup$ @yo' Not sure what definition you use, but for me binary only has two digits, which unless otherwise stated, by convention are "0" and "1". So for me it's not that something explicitly forbids "2", the problem is that its meaning is not defined and thus can't be parsed. $\endgroup$ – Oriol Aug 18 '16 at 23:31
  • 1
    $\begingroup$ @yo' In any base, you can use any character or other symbol that you choose for each value. But for any base-N, you can only ever have exactly N-1 discrete values, and hence N-1 symbols. If you choose to use the symbols "2" and "T" to represent the values for which the rest of the world uses "0" and "1", your maths will still be perfectly self-consistent. However you will be unable to communicate with the rest of the world, which some people might see as a bit of a downer. $\endgroup$ – Graham Aug 19 '16 at 9:57
  • 1
    $\begingroup$ @Graham That's not true, even the slightest. What you describe is a restriction for having unique expansions (and even then, it's disputable, given the problem with infinite expansions and the problem with negative numbers). I can happily use the binary system with digits $0,1,2$, and I don't mind that the number ten can be expressed in 5 ways: $122=202=210=1002=1010$. Seeing the digits $0,1$ as the only possibility is a very narrow view of the positional number systems. $\endgroup$ – yo' Aug 19 '16 at 16:31
  • 1
    $\begingroup$ @yo' Not true. The base defines the number of digits used. It's really that simple. You can imagine a system where each digit can count in base-3 (digits 0/1/2) but each successive place increases in powers of 2, sure. But then it's neither base-3 nor base-2. $\endgroup$ – Graham Aug 19 '16 at 16:40
4
$\begingroup$

==== better shorter answer ====

It's not true for base 2 because "2" doesn't exist in base 2.

But changing bases doesn't change what the numbers are. It only changes how we represent them. So the real question is does "2" always represent the same number that "1 + 1" represents?

And the answer to that is: Yes, so long as the base is greater than 2 so that it has the digit "2".

Any base N will have N digits: {0, 1, 2,........, (N-1)}. "0" is a "null place holder" (sort of, it's a little more complicated than that) and {1,2,3.....,(N-1)} represent the first of the natural numbers. So "2" is always the number we know as 2 if "2" exists. "3" is always the number we know as 3 if "3" exists.... "9" is always the number we know as 9 if "9" exisits. And "A" is always the number we know as 10 if "A" exists" (which it doesn't in base 10; ... or should I say "base A"?... It does exist in base 11 which has the digits {0,1,2,3,4,5,6,7,8,9,A} which represent the null place holder and the first "ten" natural numbers).

So if "2" and "1" exist. "1 + 1" and "2" always represent the number we know as 2 and the sum we know as 1 + 1.

===== third answer =====

$a + b = c$ where $a,b$ and $c$ are single digit numbers and the base of our numbering system is greater than $c$ will always be true (assuming, of course, that $a + b$ does equal $c$).

Whereas if the number system is $N \le c$ or less. $a + b = 1M$ where $M$ is the symbol for $c - N$.

The reason is because digits of a base N system are {0,1,....,N-1} and represent those natural numbers. If $a + b < N$ there will be a digit to represent it. If $a+b \ge N$ there won't be.

So for, example $4 + 3 =7$ for all bases that have a "$7$". i.e. all bases greater than 7.

So $1+1=2$ is true for all bases that have a "$2$", that is, all bases greater than $2$.

Or to put it a different way:

If $a + b = c< N$ then in base N, $a + b = a*N^0 + b*N^0 = cN^0 = c$. As $c < N$ there is a digit representing c$.

If $a + b = c \ge N; 0 \le a < N; 0 \le b < N$ then in base N $a + b = a*N^0 + b*N^0 = (a+b)*N^0 = c*N^0 = (N + (c - N))*N^0 = N^1 + (c-N)*N^0$. Note that $0 \le c-N < N$ so there is a digit $M = c-N$. So $N^1 + (c-N)*N^0 = 1*N^1 + M*N^0 = 1M$.

==== old answer ====

Um.... 1 and 2 are symbols and mean nothing by themselves.

When we do base arithmetic we share symbols for digits.

Base1 = has one digit-symbol {1}

Base2 = has two digit-symbols {0,1}

Base3 = has three {0,1,2}

Base 4 = has four {0,1,2,3}

...

And so on.

For all bases the symbol "1" means the "base single number.

For all bases that contain the symbol "2", "2" means the "number after 1".

For all bases that contain the symbol "3", "3" means the "number after 2".

And so on.

If "2" is a symbol in the base then, yes, "2" = "the number after the base number" = "the base number plus the base number" = "1+1". That will always be true.

But Base 1 and Base 2 do not have the symbol "2".

In base 2, 1+ 1 = 10.

In base 1, 1 + 1 = 11.

But for all bases > 2 which do have the symbol "2". 1 + 1 = 2.

$\endgroup$
  • 1
    $\begingroup$ It gets confusing to talk about because we start having to refer to "the number X in base n" vs "the 'real' number X in reality.... except we only call it X because we are in base 10; it's real name... well numbers don't have real names but...you know what I mean and..." $\endgroup$ – fleablood Aug 18 '16 at 19:52
  • $\begingroup$ Nonsense, nonsense, nonsense. Why wouldn't the digit $2$ exist in base $2$?! Who has forbid it?! $\endgroup$ – yo' Aug 18 '16 at 22:33
  • 1
    $\begingroup$ @yo' Is that a serious question? The digit doesn't exist in base 2 because base 2 only has the digits {0,1}. in base 2, the number 2 is represented, by definition, as 10. Base 2 represents all numbers as $a_na_{n-1}.....a_2a_1a_0 = a_n*2^n + a_{n-1}*2^{n_1} +.... + a_2*2^2+a_1*2 + a_0$ where $a_i = 0 or 1$. So the only digits are $0$ and $1$. For any base N, there are always N digits representing the integers from 0 to N-1. No base N ever has a digit for N. Thats why we don't have a digit for 10. Seriously, are you making a joke or what? $\endgroup$ – fleablood Aug 18 '16 at 23:33
  • $\begingroup$ But what prevents you from using any finite subset of $\mathbb{Z}$ as your digit set? It's for example common to use the symmetric binary expansions with digits $0, \pm1$ where the density of non-zeros is smaller which can have some applications in computing. But this doesn't mean you can't use for instance $0,2,3,-7$ as your digit set for binary representations. $\endgroup$ – yo' Aug 19 '16 at 7:27
  • 1
    $\begingroup$ @yo' Nothing prevents you from using pandas as your digit set. But it wouldn't be base 2 if you used digits other than $\{0,1\}$. $\endgroup$ – wizzwizz4 Aug 19 '16 at 9:54
2
$\begingroup$

If you put one egg in an empty basket and then put yet an other egg in the same basket, then you have 10 eggs in the basket if the base is 2 or 2 eggs in the basket if the base is greater than 2.

$\endgroup$
  • $\begingroup$ If you put one egg in a basket, and then put yet another egg in the same basket, then you have 11 eggs in the basket if the base is 1, or you will have 10 eggs in the basket if the base is 2, or you will have 2 eggs in the basket. Note: see Unary Numeral System $\endgroup$ – Will Sherwood Aug 18 '16 at 19:21
  • 1
    $\begingroup$ If you put one egg in a basket and then put yet another egg in the same basket, then you have two eggs in the basket. End of discussion. The rest is just semantics. $\endgroup$ – Wildcard Aug 18 '16 at 21:02
  • 1
    $\begingroup$ Or you have $10\overline{7}$ eggs if you wish (guess in what base ;-)) $\endgroup$ – yo' Aug 18 '16 at 22:31
2
$\begingroup$

Well, the most common definition of $2$ is "the integer that follows $1$". Since the string $a$ for any digit $a$ represents the number $a$, we get that $2=1+1$ no matter the base.

Note that this clearly shows that the definition of $2$ is "algebraic", because you can extend this not only to bases like $2$, $10$, $-5$, $1.875$ or $3+\mathrm{i}$, but also to things like $X+1$ in the ring $(\mathbb{Z}/3\mathbb{Z})[X]$ and similar beasts (yes, you can try to expand polynomials modulo $3$ as $\sum a_i (X+1)^i$ with $a_i$ in a finite set, the question is whether you'll be successful; it should work with $a_i\in\{0,1,2\}$).


Note that one can choose any base and any digit set for the base and start expressing numbers. It does not mean that every combination of a base and a digit set is a good one, since quite often not every real number has a representation, but heck, even in the standard binary system only non-negative numbers can be represented, we need the sign for the negative numbers. So yes, from the generic point of view, $2$ is fine as a digit for the binary system.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.